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Question

If z=724i, then z is equal to

A
34i
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B
34i
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C
3+4i
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D
3+4i
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Solution

The correct option is C 3+4i
Given : z=724i
Let z=a+ib, so
z2=724i(a+ib)2=724ia2b2+2i(ab)=724ia2b2=7, ab=12

Now,
(a2+b2)2=(a2b2)2+4a2b2(a2+b2)2=49+4(12)2(a2+b2)2=625a2+b2=25 (a,bR)

Using a2b2=7, a2+b2=25, we get
a2=9, b2=16a=±3, b=±4
Using ab=12, we get
a=3,b=4a=3,b=4
Hence, z=34i,3+4i

Alternate solution :
We know that,
aib=±[12(a2+b2+a)i12(a2+b2a)]
If (b>0)
Hence z=724i=7(24)i
z=±[12(72+2427)i12(72+242+7)]
z=±(34i)
z=3+4i and z=34i

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